Mathematics for the interested outsider

Last time we defined a short exact sequence in an abelian category to be an exact sequence of the form . These are the objects of a category , whose morphisms are triples of arrows , , making the two squares commute in the diagram:
The category is clearly enriched over .

As a first step into this category, we show that if and are both monic, then is monic as well. First let’s see how to do this in . We want to show that if for that . Now , but we assumed to be monic, so — . Since the top row is exact this means , and so there’s an so that . Now , but both and are monic. Thus , and as well.

That’s the normal way to proceed, and we call it a “diagram chase”. Just start with an element on the diagram at and “chase” it around the diagram. But this only works if is made up of structured sets, and we haven’t assumed that at all! We need to work a tiny bit more abstractly and look only at the arrows.

So let’s take . Now . Since is monic, we have . And thus factors uniquely through as . Now . Since and are assumed monic, , and so and is monic.

Either way, we can dualize the whole theorem to say that if and are epic, then is epic too. Together, these results are called the “short five lemma”.

Here’s another lemma, based on this diagram:
The square on the right is a pullback, and I say that if is epic then is too. On the left is the kernel of , and I say that it factors through the kernel of to make the diagram commute.

Let’s just skip the chase and go directly to the general picture. As usual, pullbacks can be defined by products and equalizers. We get the product as , which then has two arrows to : and . And we find their equalizer by taking the kernel of their difference: . Then and .

Now under the assumption on , we know that is epic. Indeed, if then . But since is assumed epic, . Supposing now that , we see that , so factors through the cokernel of , which is . That is, , and so . Since is epic, , and so . Thus is epic.

As for the other assertion, the pair of arrows and satisfy , and so there is a unique arrow by the universal property of the pullback. In particular, . On the other hand, given any other with we have , so factors uniquely through as . Then and . But since the arrow factoring through a pullback is unique we must have . So really is the kernel of , as asserted.

About this weblog

This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).

I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.